The given function is cosx−sinx 1+sin2x .
∫ cosx−sinx 1+sin2x dx (1)
Also, sin 2 x+ cos 2 x=1 and sin2x=2sinxcosx (2)
From (1) and (2),
cosx−sinx 1+sin2x = cosx−sinx ( sinx+cosx ) 2
Let,
sinx+cosx=t ( cosx−sinx )dx=dt (3)
Put (3) in (1), we get,
∫ cosx−sinx 1+sin2x dx = ∫ dt t 2 =− 1 t +c (4)
Substitute t in (4), we get
∫ cosx−sinx 1+sin2x dx = −1 sinx+cosx +c
Thus, the integral of the function cosx−sinx 1+sin2x is −1 sinx+cosx +c.